Data:- $b_f =1200 mm \\ D_f = 120 mm \\ b_w =230 mm \\ d = 400 mm \\ Ast = 6 - 16 mm \\ =1206 mm^2 \\M20, F_{e415}$

Assume $X_u \lt D_f$ [N.A lies in the flange]

$C_u=T_u\\ 0.36f_ckb_fX_u=0.87 f_yAst\\ 0.36\times 20\times1200\times X_u=0.87\times415\times1206\\ X_u=50.39 mm \lt D_f $

Assumption is correct

Now, $X_{u\space max} = 0.48d = 0.48 x 400 = 192mm$

$X_u \lt X_{u\space max} …..$ Hence under reinforced section.

$M_u=(C_{u1}\times L_{a1})\\ M_u[(0.36f_ckb_wX_{u\space max})\times (d-0.42 X_{u\space max})]\\ M_u=0.36\times20\times1200\times78.76\times(400-0.42\times50.39)\\ M_u=164.93KNm$

Type 4 :

Shear Reinforcement

factored shear force $(V_{UD} )=1.5 x v$

Nominal shear stress $(Z_{UV} ) :- $

$Z_{UV}=\dfrac {V_{UD}}{(b\times d)} \gt Z_{uc\space max} i.e, (2.8N/mm^2) \\ pt = \dfrac {100\times Ast}{bd}$

Shear stress in concrete $(Z_{uc} )$

$Z_{uc}\to $

shear load taken by concrete $(V_{uc} ):\\ V_{uc}= Z_{uc}bd$

shear load taken by minimum reinforcement $(V_{u\space min} )$

$V_{u\space min} = 0.4bd$

Assume diameter $\phi_s$

$a_{sv}$ No. of vertical legs $\times \dfrac \pi4 \phi^2_s$

If $V_{uD} \lt V_{uc}+ V_{u\space min}$

Minimum shear reinforcement is sufficient.

Spacing:-

$$S_1=\dfrac {0.87f_ya_{sv}d}{V_{u\space min}}\\ S_2=0.75d\\ S_3=300 mm $$

If $V_{UD} \gt (V_{uc}+V_{u\space min})$

Design and provide shear reinforcement

$$V_{us} = (V_{UD}-V_{uc})$$

Spacing:

$$S_1=\dfrac {0.87f_ya_{sv}d}{V_{us}}\\ S_2=0.75d\\ S_3=300 mm $$